Question

已知函数f(x)=根号2sin(2x+6π）.x∈[π/12,π/3].求f(x)的值域.

Answer 1

f(x)=(√2)sin(2x+6π)=(√2)sin(2x)

设y=2x,则：f(x)=(√2)siny

因为x∈[π/12,π/3],所以：y∈[π/6,2π/3]

显然,当y=π/2时,siny有最大值,即f(x)有最大值：f(x)|max=(√2)sin(π/2)=√2

当y=π/6时,siny有最小值,即f(x)有最小值：f(x)|min=(√2)sin(π/6)=√2/6

所以,f(x)的值域是：f(x)∈[√2/6,√2]